FFT学习笔记

快速傅里叶变换,可以将多项式相乘的时间复杂度从最简单的O(n^2)优化到O(nlgn),详细过程参考算法导论.

FFT的流程大致是:

1):构造多项式,复杂度O(n)

2):求两个多项式的DFT,复杂度O(nlgn)

3):构造多项式乘积的点值表达式,复杂度O(n)

4):求点值表达式的IDFT,复杂度O(nlgn).

下面是两道最简单的习题:

HDU 1402:点击打开链接

求两个大数乘积.

因为一个大数可以看成是一个多项式,每一位上的值都表示对应次数下的系数,因此可以用FFT加速.

本体的一个坑点就是

len = l1+l2-1;

这句代码,可能是精度问题在len更加高位的地方出现了非0值.

#include <bits/stdc++.h>
using namespace std;
#define pi acos (-1)
#define maxn 200010

struct plex {
    double x, y;
    plex (double _x = 0.0, double _y = 0.0) : x (_x), y (_y) {}
    plex operator + (const plex &a) const {
        return plex (x+a.x, y+a.y);
    }
    plex operator - (const plex &a) const {
        return plex (x-a.x, y-a.y);
    }
    plex operator * (const plex &a) const {
        return plex (x*a.x-y*a.y, x*a.y+y*a.x);
    }
};

void change (plex y[], int len) {
    if (len == 1)
        return ;
    plex a1[len], a2[len];
    for (int i = 0; i < len; i += 2) {
        a1[i/2] = y[i];
        a2[i/2] = y[i+1];
    }
    change (a1, len>>1);
    change (a2, len>>1);
    for (int i = 0; i < len/2; i++) {
        y[i] = a1[i];
        y[i+len/2] = a2[i];
    }
    return  ;
}

void fft(plex y[],int len,int on)
{
    change(y,len);
    for(int h = 2; h <= len; h <<= 1)
    {
        plex wn(cos(-on*2*pi/h),sin(-on*2*pi/h));
        for(int j = 0;j < len;j+=h)
        {
            plex w(1,0);
            for(int k = j;k < j+h/2;k++)
            {
                plex u = y[k];
                plex t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i].x /= len;
}
char a[maxn], b[maxn];
plex x1[maxn], x2[maxn];
int ans[maxn];

int main () {
    while (scanf ("%s%s", a, b) == 2) {
        int len = 2, l1 = strlen (a), l2 = strlen (b);
        while (len < l1*2 || len < l2*2)
            len <<= 1;
        for (int i = 0; i < l1; i++) {
            x1[i] = plex (a[l1-1-i]-'0', 0);
        }
        for (int i = l1; i < len; i++)
            x1[i] = plex (0, 0);
        for (int i = 0; i < l2; i++) {
            x2[i] = plex (b[l2-1-i]-'0', 0);
        }
        for (int i = l2; i < len; i++)
            x2[i] = plex (0, 0);
        fft (x1, len, 1);
        fft (x2, len, 1);
        for (int i = 0; i < len; i++)
            x1[i] = x1[i]*x2[i];
        fft (x1, len, -1);
        for (int i = 0; i < len; i++) {
            ans[i] = (int)(x1[i].x+0.5);
        }
        for (int i = 0; i < len; i++) {
            if (ans[i] >= 10) {
                ans[i+1] += ans[i]/10;
                ans[i] %= 10;
            }
        }
        len = l1+l2-1;
        while (ans[len] <= 0 && len > 0)
            len--;
        for (int i = len; i >= 0; i--) {
            printf ("%d", ans[i]);
        }
        printf ("\n");
    }
    return 0;
}

HDU 4609: 点击打开链接

题意是给出n个长度,任取3个求能组成三角形的概率.

首先记录下每个长度的数量,然后用FFT加速求出任取两个长度下的情况,这里面有重复:

首先减去两次都取同一根的情况,减完之后的结果都/2.

最后只需要所有的情况减去不能组成三角形的情况,将最初的长度序列排序后从小到大枚举下标,假设这条边是最长边,那么如果所有两条边长度小于等于这条边的情况就应该减去,这里用前缀和统计下就好了.

#include <bits/stdc++.h>
using namespace std;
#define pi acos (-1)
#define maxn 611111

struct plex {
    double x, y;
    plex (double _x = 0.0, double _y = 0.0) : x (_x), y (_y) {}
    plex operator + (const plex &a) const {
        return plex (x+a.x, y+a.y);
    }
    plex operator - (const plex &a) const {
        return plex (x-a.x, y-a.y);
    }
    plex operator * (const plex &a) const {
        return plex (x*a.x-y*a.y, x*a.y+y*a.x);
    }
};

void change (plex y[], int len) {
    if (len == 1)
        return ;
    plex a1[len/2], a2[len/2];
    for (int i = 0; i < len; i += 2) {
        a1[i/2] = y[i];
        a2[i/2] = y[i+1];
    }
    change (a1, len>>1);
    change (a2, len>>1);
    for (int i = 0; i < len/2; i++) {
        y[i] = a1[i];
        y[i+len/2] = a2[i];
    }
    return  ;
}

void fft(plex y[],int len,int on)
{
    change(y,len);
    for(int h = 2; h <= len; h <<= 1)
    {
        plex wn(cos(on*2*pi/h),sin(on*2*pi/h));
        for(int j = 0;j < len;j+=h)
        {
            plex w(1,0);
            for(int k = j;k < j+h/2;k++)
            {
                plex u = y[k];
                plex t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i].x /= len;
}
long long num[maxn], sum[maxn];
int a[maxn];
plex x[maxn];
long long n;

int main () {
    //freopen ("in.txt", "r", stdin);
    int t;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%lld", &n);
        long long Max = 0;
        memset (num, 0, sizeof num);
        for (int i = 1; i <= n; i++) {
            scanf ("%d", &a[i]);
            num[a[i]]++;
            Max = max (Max, (long long)a[i]);
        }
        Max++;
        int len = 2;
        while (len < Max*2)
            len <<= 1;
        for (int i = 0; i < len; i++) {
            x[i] = plex (num[i], 0);
        }
        fft (x, len, 1);
        for (int i = 0; i < len; i++) {
            x[i] = x[i]*x[i];
        }
        fft (x, len, -1);
        for (int i = 0; i < len; i++) {
            num[i] = (long long) (x[i].x+0.5);
        }
        for (int i = 1; i <= n; i++) {//两次取同一个
            num[a[i]+a[i]]--;
        }
        for (int i = 0; i < len; i++) {//重复计算
            num[i] /= 2;
        }
        sum[0] = 0;
        for (int i = 1; i < len; i++) {
            sum[i] = sum[i-1]+num[i];
        }
        sort (a+1, a+1+n);
        long long tot = n*(n-1)*(n-2)/6, ans = tot;
        for (int i = 3; i <= n; i++) {
            ans -= sum[a[i]];
        }
        printf ("%.7f\n", ans*1.0/tot);
    }
    return 0;
}